Two particles are fixed to an x axis: particle 1 of charge -1.00 x 10-7 c is at the origin and particle 2 of charge +1.00 x 10-7 c is at x = 17.1 cm. midway between the particles, what is the magnitude of the net electric field?

The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint. 

The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = -7.041 N/C

Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*7.041 N/C = 14.082 N/C

3x+9+4x+x=8x+9that's the answer for your question.